Results 1 to 3 of 3

Math Help - Limit

  1. #1
    Super Member
    Joined
    Jun 2008
    Posts
    829

    Limit

    Consider a function f that |f(x) - Ax| \leq e^{-x}, for any number A \in R. Calculate:

    lim_x{ \to +\infty} \frac{f(x) - f(1)}{x-1}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Hi

    \frac{|f(x) - f(1)|}{|x-1|} = \frac{|(f(x) - Ax) + (Ax - A) + (A - f(1))|}{|x-1|}

    \frac{|f(x) - f(1)|}{|x-1|} \leq \frac{|f(x) - Ax| + |Ax - A| + |A - f(1)|}{|x-1|}

    Spoiler:

    \frac{|f(x) - f(1)|}{|x-1|} \leq \frac{|f(x) - Ax|}{|x-1|} + \frac{|Ax - A|}{|x-1|} + \frac{|A - f(1)|}{|x-1|}

    \frac{|f(x) - f(1)|}{|x-1|} \leq \frac{e^{-x}}{|x-1|} + |A| + \frac{e^{-1}}{|x-1|}

    And

    \lim_{x \rightarrow +\infty} \left(\frac{e^{-x}}{|x-1|} + |A| + \frac{e^{-1}}{|x-1|}\right) = |A|

    \forall A \in \mathbb{R} ~ \lim_{x \rightarrow +\infty}\frac{|f(x) - f(1)|}{|x-1|} \leq |A|

    Therefore \lim_{x \rightarrow +\infty}\frac{|f(x) - f(1)|}{|x-1|} = 0


    It would be better if someone could confirm !

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2008
    Posts
    829
    Thank you men
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 12
    Last Post: August 26th 2010, 11:59 AM
  2. Replies: 1
    Last Post: August 8th 2010, 12:29 PM
  3. Replies: 1
    Last Post: February 5th 2010, 04:33 AM
  4. Replies: 16
    Last Post: November 15th 2009, 05:18 PM
  5. Limit, Limit Superior, and Limit Inferior of a function
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 3rd 2009, 06:05 PM

Search Tags


/mathhelpforum @mathhelpforum